leetcode之Intersection of Two Linked Lists

最新推荐文章于 2025-07-14 21:00:27 发布

howtogetout

最新推荐文章于 2025-07-14 21:00:27 发布

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本文链接：https://blog.youkuaiyun.com/howtogetout/article/details/50483289

本文详细介绍了如何通过计算两个链表长度并利用指针移动来找出它们的交集节点，包括代码实现和核心逻辑解释。

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这道题据说是个经典题啦，可惜我这种半路出身的一开始还真是没想到。解题思路是先算出来2个列表的长度，相减就是长的多出来的部分。然后切掉，然后2边一起开始走，如果是有连接点，则一定能相遇，否则走到尾没相遇的话就返回None。代码如下：

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        if not headA:
            return None
        if not headB:
            return None
        lenofheadA = 0
        lenofheadB = 0
        headB1 = headB
        headA1 = headA
        while headB1:
            lenofheadB += 1
            headB1 = headB1.next
        while headA1:
            lenofheadA += 1
            headA1 = headA1.next
        deviation = abs(lenofheadA - lenofheadB)
        if lenofheadA >= lenofheadB:
            for i in range(deviation):
                headA = headA.next
            NumofStep = 0
            while NumofStep < lenofheadB:
                if headA == headB:
                    return headB
                else:
                    headB = headB.next
                    headA = headA.next
                NumofStep += 1
            else:
                return None
        else:
            for i in range(deviation):
                headB = headB.next
            NumofStep = 0
            while True:
                if headA == headB:
                    return headB
                else:
                    headB = headB.next
                    headA = headA.next
                NumofStep += 1
            else:
                return None